Question

Помогите пожалуйста Срочно!

Answer 1

Ответ:

1

а)

$log_{3}(x) = - 2 \\ x = {3}^{ - 2} \\ x = \frac{1}{9}$

б)

$log_{x}(27) = - 3 \\ {x}^{ - 3} = 27 \\ x = {27}^{- \frac{1}{3} } \\ x = \frac{1}{3}$

в)

$log_{2}(32) = x \\ {2}^{x} = 32 \\ x = 5$

2.

а)

$log_{12}(3) + log_{12}(4) = log_{12}(3 \times 4) = \\ = log_{12}(12) = 1$

б)

$\frac{ log_{2}(5) }{ log_{2}(25) } = log_{25}(5) = log_{ {5}^{2} }(5) = \frac{1}{2}$

в)

${( \frac{1}{49}) }^{ log_{ \frac{1}{7} }(5) } = {( \frac{1}{7}) }^{2 log_{ \frac{1}{7} }(5) } = \\ = {( \frac{1}{7}) }^{ log_{ \frac{1}{7} }( {5}^{2} ) } = {5}^{2} = 25$

г)

$log_{0.1}(4) - log_{10}(25) = log_{ {10}^{ - 1} }(4) - log_{10}(25) = \\ = - log_{10}(4) - log_{10}(25) = - ( log_{10}(4) + log_{10}(25) ) = \\ = - log_{10}(100) = - 2$

3.

а)

$log_{0.4}(2x - 5) = - 2 \\ 2x - 5 = {(0.4)}^{ - 2} \\ 2x - 5 = {( \frac{1}{5} )}^{ - 2} \\ 2x - 5 = 25 \\ 2x = 30 \\ x = 15$

б)

$log_{3}(2 {x}^{2} + 4x - 7) = log_{3}( {x}^{2} - x - 1 ) \\ 2 {x}^{2} + 4x - 7 = {x}^{2} - x - 1 \\ {x}^{2} + 5x - 6 = 0 \\ d = 25 + 24 = 49 \\ x1 = \frac{ - 5 + 7}{2} = 1 \\ x2 = - 6$

Проверка:

$x1 = 1 \\ \\ log_{3}(2 + 4 - 7) = log_{3}(1 - 1 - 1)$

здесь в скобках получаются числа меньше 0, значит корень не подходит.

$x2 = - 6 \\ \\ log_{3}(2 \times 36 - 24 - 7) = log_{3}(36 + 6 - 1) \\ log_{3}(72 - 31) = log_{3}(36 + 5) \\ log_{3}(41) = log_{3}(41)$

подходит.

Ответ: - 6

в)

$2 log_{3}( - x) = 3 + log_{3}( \frac{x}{9} + 1) \\ log_{3}( {( - x)}^{2} ) = log_{3}(27) + log_{3}( \frac{x}{9} + 1 ) \\ x^{2} = 27( \frac{x}{9} + 1) \\ {x}^{2} - 3x - 27 = 0 \\ d = 9 + 108 = 117 = 9 \times 13 \\ x1 = \frac{3 + 3 \sqrt{13} }{2} \\ x2 = \frac{3 - 3 \sqrt{13} }{2}$

ОДЗ:

$- x > 0 \\ \frac{x}{9} + 1 > 0 \\ \\ x < 0 \\ x > - 9$

подходит только второй корень

Ответ:

$x = \frac{3 - 3 \sqrt{13} }{2}$

г)

$log_{3}^{2} ( \frac{x}{81} ) + 4 log_{3}(x) - 61 = 0 \\ (log_{3}(x) - log_{3}(81)) ^{2} + 4 log_{3}(x) - 61 = 0 \\( log_{ 3} (x) - 4)^{2} + 4 log_{3}(x) - 61 = 0 \\ log_{3} ^{2} (x ) - 8 log_{3}(x) + 16 + 4 log_{3}(x) - 61 = 0 \\ log_{3}^{2} (x) - 4 log_{3}(x) - 45 = 0 \\ \\ log_{3}(x) = t \\ {t}^{2} - 4t - 45 = 0 \\ d = 16 + 180 = 194 \\ t1 = \frac{4 + 14}{2} = 9 \\ t2 = - 5 \\ \\ log_{3}(x) = 9 \\ x1 = {3}^{9} \\ x1 = 19683 \\ \\ log_{3}(x) = - 5 \\ x2 = \frac{1}{243}$

оба корня подходят.